Question

An alternating voltage is given as, \(E=100\sqrt{2}\sin 100t\)
This voltage E is applied to a capacitor of \(1\ \mu \text{F}\). The current reading in an a.c. ammeter in mA is

• A. \(10\)
• B. \(20\)
• C. \(40\)
• D. \(80\)

Hint:

An AC ammeter reads the rms current. Find the capacitive reactance from the angular frequency and capacitance, then divide the rms voltage by it.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Given \(E=100\sqrt{2}\sin100t\), so \(V_{0}=100\sqrt{2}\ \text{V}\) and \(\omega=100\ \text{rad/s}\).

Step 2: Reactance
\(X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{100\times10^{-6}}=10^{4}\ \Omega\)

Step 3: Current
\(V_{rms}=\dfrac{V_{0}}{\sqrt{2}}=100\ \text{V}\)
\(I_{rms}=\dfrac{100}{10^{4}}=10^{-2}\ \text{A}=10\ \text{mA}\)

Final Answer:
The ammeter reads 10 mA. \[ \boxed{\text{(A) }10} \]